How do I solve this equation in the field of complex numbers?: $$|z|^2 - z|z| + z = 0 $$ My solutions are: $$z_1 = 0$$ $$z_2 = -1$$
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1$\begingroup$ $-1$ is not a solution as you would get $1+1-1\not =0$ $\endgroup$ – Henry Nov 21 '18 at 11:20
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$$z=\dfrac{|z|^2}{|z|-1}$$ which is real
If $z>0,|z|=+z$ $$0=z^2-z^2+z\iff z=0$$ which is untenable
If $z\le0,|z|=-z$ $$0=z^2+z^2+z=z(2z+1)$$
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$\begingroup$ Your first step only works when $|z| \ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.) $\endgroup$ – ruakh Nov 21 '18 at 23:29
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Observe that $z=0$ is a solution of the equation
$$|z|^2 - z|z| + z = 0.$$
( $z=-1$ is not a solution !)
Therefore let $z \ne 0$ be a further solution of this equation. We get , since $|z|^2=z \overline{z}:$
$\overline{z}-|z|+1=0$. This gives $\overline{z}=|z|-1 \in \mathbb R$.Hence
$z=|z|-1 \in \mathbb R$.
If $z>0$ , we have $z=z-1$, which is impossible. Hence $z<0$ and then $z=-1/2$.
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WLOG $z=r(\cos t+i\sin t)$ where $r>0,t$ are real
$$r(r-r(\cos t+i\sin t)+(\cos t+i\sin t))=0$$
If $r\ne0,$ equating the real & the imaginary parts
$r-r\cos t+\cos t=0=(1-r)\sin t$
Case $\#1:$
If $r=1,1=0$ which is untenable
If $\sin t=0,$
Case $\#2A:\cos t=1,r=0$ which is untenable
Case $\#2A:\cos t=-1,r+r-1=0\iff r=?$
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You may also proceed as follows:
- Rewrite the equation to $$|z|^2 - z|z| + z = 0 \Leftrightarrow \boxed{|z|^2 = z(|z|-1)}$$
- Therefore, $\color{blue}{z}$ must be $\color{blue}{\mbox{real}}$ and so we have $\color{blue}{|z|^2 = z^2}$.
- Noting the solution $\boxed{z = 0}$ we get $$|z|^2 = z(|z|-1) \stackrel{z \in \mathbb{R}, z \neq 0}{\Leftrightarrow}z = |z|-1 \Rightarrow \boxed{z = -\frac{1}{2}}$$
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Let $r = e^{i\theta}$.
We get $r^2 - re^{i\theta}\cdot r + re^{i\theta} = 0$.
Factorising, we get $r = 0$, giving $z = 0$ as one solution or:
$r - re^{i\theta} + e^{i\theta} = 0$
$e^{i\theta} = \frac{r}{r-1}$
From the last equation, the magnitude of the LHS is equal to one. From the RHS, $e^{i\theta} \in \mathbb{R}$. Hence $e^{i\theta} = \pm 1$.
$\frac{r}{r-1} = 1$ gives no solution, but $\frac{r}{r-1} = -1 \implies r = \frac 12$. Since $e^{i\theta} = -1$ that gives $z = -\frac 12$.
Therefore the two solutions are $z = 0$ and $z = -\frac 12$.
